GCSELevel 6-7Cambridge iGCSE

Solving Quadratic Equations by Factorisation Revision

Quadratic equations are equations that take the form, or can be rearranged to the form,

$ax^{2}+bx+c=0$

These can be solved by factorising the $ax^{2}+bx+c$ part of the equation into two brackets, then setting each bracket to $0$ to derive two solutions, called roots.

Note: Not all quadratics can be factorised, so not all quadratics can be solved with this method.

Make sure you are happy with the following topics before continuing:

How to Solve a Quadratic Equation by Factorisation

The general steps to solve a quadratic by factorisation are as follows:

Step 1: If necessary, rearrange the quadratic into the form $ax^{2}+bx+c=0$

Step 2: Factorise the quadratic into two brackets $(px+q)(rx+s)=0$

Step 3: Get two solutions by solving the two linear equations $px+q=0$ and $rx+s=0$

Example: Solve by factorisation $x^{2}+4x+3=0$

Step 1: This quadratic is already in the form we want, so we don’t need to rearrange.

Step 2: Factorise the quadratic: $x^{2}+4x+3=(x+3)(x+1)=0$

Step 3: Set both brackets to $0$ and solve:

$x+3=0\rightarrow x=-3$ and $x+1=0\rightarrow x=-1$

So our two solutions are $x=-3$ and $x=-1$

Tip: For quadratics with $a=1$, we can read the solutions off from the factorisation, as the roots of $(x+p)(x+q)=0$ are $-p$ and $-q$. For quadratics with $a\neq1$ it is more complicated.

Example: Solve by factorisation $2x^{2}-11x=-5$

Step 1: Rearrange by adding $5$ to both sides: $2x^{2}-11x+5=0$

Step 2: Factorise the quadratic: $(2x-1)(x-5)=0$

Step 3: Set both brackets to $0$ and solve:

$2x-1=0\rightarrow x=\dfrac{1}{2}$ and $x-5=0\rightarrow x=5$

So our two solutions are $x=\dfrac{1}{2}$ and $x=5$

Note: The reason this method to solve quadratics works is that, once factorised, the quadratic is two brackets that multiply together to give $0$. This can only happen when one of the brackets is $0$, so we can get solutions by assuming that is the case.

Level 6-7GCSECambridge iGCSE

We can use factorisation to help us sketch a quadratic graph. This is because the roots of the quadratic equation is where the graph crosses the $x$-axis, so we can factorise to find the roots and use these to help determine the correct shape of the graph.

Example: Sketch the graph $y=x^{2}-2x-3$

Step 1: For a graph we want it to be in the form $y=ax^{2}+bx+c$, which we already have. We then set $y=0$.

$x^{2}-2x-3=0$

Step 2: Factorise the quadratic: $(x-3)(x+1)=0$

Step 3: Set both brackets to $0$ and solve:

$x-3=0\rightarrow x=3$ and $x+1=0\rightarrow x=-1$

Alternatively we can use the tip from earlier to read off the roots from the brackets as $x=3$ and $x=-1$

Step 4: These roots are the $x$-intercepts of the graph – the places where the graph crosses the $x$-axis. So we know the graph crosses the points $(3,0)$ and $(-1,0)$.

Step 5: Find the $y$-intercept. For a graph of the form $y=ax^{2}+bx+c$, this is just the number on the end, $c$. In this example, it is $-3$, so the graph also goes through $(0,-3)$.

Step 6: Use the points we have found to sketch the graph. The graph on the right was drawn by a computer, a sketch will not need to look as accurate as this. Instead, a sketch only has to be the right shape and cross the axes at the points we found.

Level 6-7GCSECambridge iGCSE

Example 1: Solving a Quadratic Equation by Factorisation ($a=1$)

Solve by factorisation $x^{2}=7x-12$

[2 marks]

Step 1: Rearrange by subtracting $7x-12$ from both sides:

$x^{2}-7x+12=0$

$x^{2}-7x+12=(x-4)(x-3)=0$

Step 3: Find the solutions by setting both brackets equal to $0$:

$x-4=0\rightarrow x=4$ and $x-3=0\rightarrow x=3$

Alternatively, since $a=1$, the solutions $x=4$ and $x=3$ can be read off from the factorisation.

Level 6-7GCSECambridge iGCSE

Example 2: Solving a Quadratic Equation by Factorisation ($a\neq1$)

Solve by factorisation $5x^{2}+54x-11=0$

[3 marks]

Step 1: Rearrange – in this case it is not necessary to do so as the equation is already in the form we want.

$5x^{2}+54x-11=(5x-1)(x+11)=0$

Step 3: Set both brackets to $0$ and solve:

$5x-1=0\rightarrow x=\dfrac{1}{5}$ and $x+11=0\rightarrow x=-11$

So the solutions are:

$x=\dfrac{1}{5}$ and $x=-11$

Level 6-7GCSECambridge iGCSE

Example 3: Factorisation and Quadratic Graphs

Sketch the graph $y=x^{2}-x-6$

[4 marks]

Step 1: This quadratic graph is already in the form we are looking for. The quadratic we want to solve is $x^{2}-x-6=0$

$x^{2}-x-6=(x-3)(x+2)=0$

Step 3: Solve the equation by setting both brackets to $0$:

$x-3=0\rightarrow x=3$ and $x+2=0\rightarrow x=-2$

Alternatively, the solutions $x=3$ and $x=-2$ can be read off from the factorisation.

Step 4: These roots are the $x$-intercepts of the graph, so we know the graph passes through the points $(3,0)$ and $(-2,0)$

Step 5: The $y$-intercept is the number on the end of the quadratic, in this case it is $-6$, so the graph passes through $(0,-6)$

Step 6: Sketch the graph, making sure it is the right shape and it crosses the axes at the right points.

Level 6-7GCSECambridge iGCSE

Solving Quadratic Equations by Factorisation Example Questions

Step 1: Rearrange if necessary – this equation is already in the correct form.

$x^{2}-5x+4=(x-4)(x-1)=0$

Step 3: Set both brackets to $0$ and solve:

$x-4=0\rightarrow x=4$ and $x-1=0\rightarrow x=1$

Alternatively, we can read the solutions $x=4$ and $x=1$ off from the factorisation.

Step 1: Rearrange the equation. In this case we need to subtract $21x+72$ from both sides:

$x^{2}-21x-72=0$

$x^{2}-21x-72=(x-24)(x+3)=0$

Step 3: Set both brackets to $0$ and solve:

$x-24=0\rightarrow x=24$ and $x+3=0\rightarrow x=-3$

Alternatively, we can read the solutions $x=24$ and $x=-3$ off from the factorisation.

Step 1: Rearrange if necessary – this equation is already in the correct form.

$2x^{2}+11x+12=(2x+3)(x+4)=0$

Step 3: Set both brackets to $0$ and solve:

$2x+3=0\rightarrow x=-\dfrac{3}{2}$ and $x+4=0\rightarrow x=-4$

So the solutions are:

$x=-\dfrac{3}{2}$ and $x=-4$

Step 1: Rearrange the equation. First multiply by $3$:

$3x^{2}=21+2x$

Now subtract $21+2x$ from both sides:

$3x^{2}-2x+21=0$

$3x^{2}-2x-21=(3x+7)(x-3)=0$

Step 3: Set both brackets to $0$ and solve:

$3x+7=0\rightarrow x=-\dfrac{7}{3}$ and $x-3=0\rightarrow x=3$

So the solutions are:

$x=-\dfrac{7}{3}$ and $x=3$

Step 1: This quadratic graph is already in the form we are looking for. The quadratic we want to solve is $25x^{2}-25x+6=0$

$25x^{2}-25x-6=(5x-2)(5x-3)=0$

Step 3: Solve the equation by setting both brackets to $0$:

$5x-2=0\rightarrow x=0.4$ and $5x-3=0\rightarrow x=0.6$

Step 4: These roots are the $x$-intercepts of the graph, so we know the graph passes through the points $(0.4,0)$ and $(0.6,0)$

Step 5: The $y$-intercept is the number on the end of the quadratic, in this case it is $6$, so the graph passes through $(0,6)$

Step 6: Sketch the graph, making sure it is the right shape and it crosses the axes at the right points.

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